Coconut Piles

Date: 10/12/98 at 20:47:02
From: Fallon Stillman
Subject: Algebra Puzzle Problem
Here's the problem:
Gilligan and his buddies were stranded on a desert island. Gilligan,
the professor, Ginger, Mary Anne, the Skipper, and Fred the monkey were
gathering coconuts. One evening they all rounded up all the coconuts
they could find and put them in one large pile. Being exhausted from
working so hard they decided to wait and divide them up evenly in the
morning.
During the night Gilligan awoke and separated the nuts into five equal
piles, with one left over which he gave to Fred the monkey. Gilligan
took one pile, hid it, pushed the other four back together, and went
back to his hut. He was followed by Ginger, Mary Ann, the Professor,
and the Skipper, each dividing them up equally with one remaining nut
going to Fred. The next morning the remaining nuts were divided up
equally with one remaining going to Fred. What is the least number of
coconuts they could have started with?

Date: 10/13/98 at 13:17:58
From: Doctor Rob
Subject: Re: Algebra Puzzle Problem
Hello Fallon,
See the following web page for a related, but not identical(!)
problem:
http://mathforum.org/dr.math/problems/koestler12.18.97.html
The difference is that in this problem, the monkey gets his coconut
before the five-way split, and in the problem on the other Web page, he
gets it after. Subtle, but important. The equations for your problem
are these.
Let a be the number of coconuts to start with. After the first person
and the monkey take their coconuts, the number left is b = (4/5)*(a-1).
After the second person and the monkey take their coconuts, the number
left is c = (4/5)*(b-1). The third person leaves d = (4/5)*(c-1)
coconuts. The fourth person leaves e = (4/5)*(d-1) coconuts. The fifth
person leaves f = (4/5)*(e-1) coconuts. At the end, f = 5*g + 1, where
g is the number of coconuts each person gets in the morning.
Now when you perform the chain substitution to find the equation
relating a and g, you get:
4*(4*[4*(4*[4*(a-1)/5-1]/5-1)/5-1]/5-1)/5 = 5*g + 1
1024*a - 15625*g = 11529
For this you have to find the smallest positive integer solution. You
can follow the model in the cited problem to figure this out. Note
that g has to be odd, so g = 2*h+1, and:
512*a - 15625*h = 13577
Now h must be odd, so h = 2*i + 1, and:
256*a - 15625*i = 14601
You can continue in this way, reducing one or the other of the
coefficients on the lefthand side until one of them is reduced to 1.
Then that variable can be expressed as an integer times the remaining
variable (say z), plus another integer. Undoing all the substitutions
will express everything as an integer function of z, and the value of
z which gives the smallest positive a is the one you want. Then you can
compute that smallest positive value of a.
There is another way, using the Extended Euclidean Algorithm. You can
use that to find that:
1024*10849 - 15625*711 = 1
Now multiply by 11529, and reduce the a-value modulo 15625 to get the
answer.
- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/